Question

a bacteria culture is growing at a rate proportional to its size. the count in the bacteria culture was 500 after 21 minutes and 13000 after 30 minutes.
(a) what was the initial size of the culture?
0.25
(b) find the doubling time.
1.9149 min
(c) find the population after 90 minutes.
3.08*10^13
(d) when will the population reach 12000?
29.76 min

Explanation:

Step1: Set up the exponential growth model

The general form of exponential growth is \(P(t)=P_0e^{kt}\), where \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, and \(k\) is the growth constant.
We know that \(P(21) = 500\), so \(500=P_0e^{21k}\), and \(P(30)=13000\), so \(13000 = P_0e^{30k}\).

Step2: Solve for \(k\)

Divide the second equation by the first equation: \(\frac{13000}{500}=\frac{P_0e^{30k}}{P_0e^{21k}}\).
Simplify to get \(26 = e^{9k}\).
Take the natural logarithm of both sides: \(\ln(26)=9k\), so \(k=\frac{\ln(26)}{9}\approx0.33\).

Step3: Solve for \(P_0\)

Substitute \(k\) into \(500 = P_0e^{21k}\).
\(P_0=\frac{500}{e^{21\times\frac{\ln(26)}{9}}}\).
Simplify \(e^{21\times\frac{\ln(26)}{9}}=e^{\frac{7\ln(26)}{3}}=(26)^{\frac{7}{3}}\).
\(P_0=\frac{500}{(26)^{\frac{7}{3}}}\approx 2\).

Step4: Find the doubling - time formula

For exponential growth \(P(t)=P_0e^{kt}\), when \(P(t) = 2P_0\), \(2P_0=P_0e^{kt}\).
Cancel \(P_0\) and get \(2=e^{kt}\), then \(t=\frac{\ln(2)}{k}\).
Since \(k = \frac{\ln(26)}{9}\), \(t=\frac{9\ln(2)}{\ln(26)}\approx1.9149\).

Step5: Find the population at \(t = 90\)

\(P(90)=P_0e^{90k}\), with \(P_0\approx2\) and \(k=\frac{\ln(26)}{9}\).
\(P(90)=2e^{90\times\frac{\ln(26)}{9}}=2\times(26)^{10}\approx3.08\times 10^{13}\).

Step6: Find \(t\) when \(P(t)=12000\)

\(12000 = P_0e^{kt}\), substitute \(P_0\approx2\) and \(k=\frac{\ln(26)}{9}\).
\(6000=e^{\frac{\ln(26)}{9}t}\).
Take the natural logarithm: \(\ln(6000)=\frac{\ln(26)}{9}t\).
\(t=\frac{9\ln(6000)}{\ln(26)}\approx29.76\).

Answer:

(a) The initial size of the culture is approximately \(2\).
(b) The doubling time is approximately \(1.9149\) minutes.
(c) The population after \(90\) minutes is approximately \(3.08\times 10^{13}\).
(d) The population will reach \(12000\) at approximately \(29.76\) minutes.